Codeforces_GYM_100741 A
http://codeforces.com/gym/100741/problem/A
A. Queries
0.25 seconds
64 megabytes
standard input
standard output
Mathematicians are interesting (sometimes, I would say, even crazy) people. For example, my friend, a mathematician, thinks that it is very fun to play with a sequence of integer numbers. He writes the sequence in a row. If he wants he increases one number of the sequence, sometimes it is more interesting to decrease it (do you know why?..) And he likes to add the numbers in the interval [l;r]. But showing that he is really cool he adds only numbers which are equal some mod (modulo m).
Guess what he asked me, when he knew that I am a programmer? Yep, indeed, he asked me to write a program which could process these queries (n is the length of the sequence):
- + p r It increases the number with index p by r. (
,
)
You have to output the number after the increase.
- - p r It decreases the number with index p by r. (
,
) You must not decrease the number if it would become negative.
You have to output the number after the decrease.
- s l r mod You have to output the sum of numbers in the interval
which are equal mod (modulo m). (
) (
)
The first line of each test case contains the number of elements of the sequence n and the number m. (1 ≤ n ≤ 10000) (1 ≤ m ≤ 10)
The second line contains n initial numbers of the sequence. (0 ≤ number ≤ 1000000000)
The third line of each test case contains the number of queries q (1 ≤ q ≤ 10000).
The following q lines contains the queries (one query per line).
Output q lines - the answers to the queries.
3 4
1 2 3
3
s 1 3 2
+ 2 1
- 1 2
2
3
1
m个 树状数组记录a[i]%m,得值。
#define LL long long
#include <iostream> using namespace std; const int MAXN();
const int N(+);
LL n,m,a[N],q,u,v,w; struct Tree
{
LL mm;
LL val[N];
#define lowbit(x) (x&(-x))
void Update(int now,int x)
{
for(;now<=mm;now+=lowbit(now)) val[now]+=x;
}
LL Query(int x)
{
LL ret=;
for(;x;x-=lowbit(x)) ret+=val[x];
return ret;
}
}tree[]; int main()
{
cin>>n>>m;
for(LL i=;i<m;i++) tree[i].mm=n;
for(LL i=;i<=n;i++)
{
cin>>a[i];
tree[a[i]%m].Update(i,a[i]);
}
cin>>q;
for(char ch[];q--;)
{
cin>>ch>>u>>v;
if(ch[]=='+')
{
tree[a[u]%m].Update(u,-a[u]);
a[u]+=v;
tree[a[u]%m].Update(u,a[u]);
cout<<a[u]<<endl;
} else
if(ch[]=='-')
{
if(a[u]<v) cout<<a[u]<<endl;
else
{
tree[a[u]%m].Update(u,-a[u]);
a[u]-=v;
tree[a[u]%m].Update(u,a[u]);
cout<<a[u]<<endl;
}
} else
if(ch[]=='s')
{
cin>>w;
cout<<tree[w].Query(v)-tree[w].Query(u-)<<endl;
}
}
return ;
}
Codeforces_GYM_100741 A的更多相关文章
随机推荐
- DEDECMS教程:列表页缩略图随机调用
如果用过DEDECMS的朋友应该都知道,有些模板列表页面需要用到缩略图,调用内容中的缩略图可以使用系统自带的脚本调用第一张图片.但是,并不是我们所有的内容里都有图片,有时候第一张图片也不一定是适合尺寸 ...
- ES6学习笔记(三)字符串的扩展
ES6 加强了对 Unicode 的支持,并且扩展了字符串对象. 1.字符的Unicode表示法 JavaScript 允许采用\uxxxx形式表示一个字符,其中xxxx表示字符的 Unicode 码 ...
- tensorflow学习之路----保存和提取数据
#保存数据注意他只能保存变量,不能保存神经网络的框架.#保存数据的作用:保存权重有利于下一次的训练,或者可以用这个数据进行识别#np.arange():arange函数用于创建等差数组,使用频率非常高 ...
- ldd---程序所需要的动态链接库
ldd本身不是一个程序,而仅是一个shell脚本:ldd可以列出一个程序所需要得动态链接库(so) [root@xiaolizi ~ ]$ ldd /usr/bin/ls linux-vdso.so. ...
- Yum数据库错误
使用yum时提示数据库错误: /var/lib/rpm... open... db4 error from db->close:... 解决办法: 1.删除/var/lib/rpm目录下的__d ...
- 9.9递归和动态规划(六)——打印n对括号的所有有效组合(即左右括号正确配对)
/** * 功能:打印n对括号的所有有效组合(即左右括号正确配对). */ 两种方法: 方法一: /** * 思路:在括号的最前面或者原有的每对括号中面插入一对括号. 至于其它任何位置.比方字符串的 ...
- 下载eclipse详细步骤
先登陆eclipse的官网 然后点击红色箭头进行选择你电脑是32还是64位的 根据自己的需求下载 然后点击下载 这里下载的是安装包,你要进行压缩.安装时一定要好相应的jdk要不然就会报错 这上面的错误 ...
- Lucene 定义
福利 => 每天都推送 欢迎大家,关注微信扫码并加入我的4个微信公众号: 大数据躺过的坑 Java从入门到架构师 人工智能躺过的坑 Java全栈大联盟 ...
- Linux系统下到哪儿寻找硬件错误
Linux系统下到哪儿寻找硬件错误 当linux系统出现故障的时候,作为管理员首先要定位错误,现在linux有许多工具都能帮助用户寻找错误,要学会利用他们确定问题.这些工具包括dmesg. ...
- matplotlib散点图笔记
定义: 由一组不连续的点完成的图形 散点图: 包含正相关性,负相关性和不相关性. 散点图生成函数: plt.scatter(x,y) 演示代码如下: import numpy as np import ...